I'm reading Filmore's paper On sums of projections. In this paper Filmore made the following remark (on page 151, last paragraph):
By elementary spectral theory any self-adjoint operator with spectrum is $[-2, 2]$ can be written as the sum of two self-adjoint operator each having spectrum in $[-2, 2]$ and large null space (i.e. dimension of the null space is equal to the dim of the underlying separable Hilbert space).
The version of spectral theorem I know is:
Spectral Theorem. For any normal operator $A$ there is a projection valued measure $E$ on $\sigma(A)$ such that $$A = \int_{\sigma(A)} z dE(z)$$
If we choose $P=E([-2, 0))$ and $Q=E([0, 2))$ then $A = AP + AQ$ but $AP$ may not have large null space!
Any help is appreciated. Thanks.
If $\sigma(A)$ is a finite set then there is an eigenvalue $\lambda$ such that $\mathcal{H}_\lambda:=\dim\ker (A-\lambda I)=\infty.$ We can decompose $\mathcal{H}_\lambda=\mathcal{H}_{\lambda,1}+\mathcal{H}_{\lambda,2}$ such that $\mathcal{H}_{\lambda,1}\perp \mathcal{H}_{\lambda,2}$ and $\dim\mathcal{H}_{\lambda_1}=\dim\mathcal{H}_{\lambda_2}=\infty.$ Then $$A=AP_{\mathcal{H}_{\lambda,1}}+A(I-P_{\mathcal{H}_{\lambda,1}})$$ have large null spaces, as the first summand vanishes on $\mathcal{H}_{\lambda_2},$ while the second on $\mathcal{H}_{\lambda_1}.$
Assume $\sigma(A)$ is infinite. Then it contains a strictly monotonic sequence sequence $\lambda_n.$ Assume $\lambda_n$ is strictly increasing. Then for $\delta_n={1\over 2}\min\{\lambda_{n+1}-\lambda_n,\lambda_n-\lambda_{n-1}\}$ the projections $P_n:=E_{(\lambda_n-\delta_n,\lambda_n+\delta_n)}$ are nontrivial and mutually orthogonal. Let $$P=\sum_{n=1}^\infty P_{2n}$$ Then $P$ is an orthogonal projection commuting with $A.$ Let $$A=AP+A(I-P)$$ Then both operators are self-adjoint, with norm less than or equal $2.$ They admit large null spaces. Namely the first summand vanishes on the range of $\sum_{n=1}^\infty P_{2n+1}$ while the second on the range of $\sum_{n=1}^\infty P_{2n}.$
When the sequence $\lambda_n$ is strictly decreasing, we consider $-A,$ decompose it according to the method described above, and multiply the decomposition by $-1.$